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# The area of the right triangle ABC is 4 times greater than

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The area of the right triangle ABC is 4 times greater than [#permalink]

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05 Feb 2012, 05:07
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KLM Triangle.GIF [ 2.03 KiB | Viewed 9281 times ]
The area of the right triangle ABC is 4 times greater than the area of the right triangle KLM. If the hypotenuse KL is 10 inches, what is the length of the hypotenuse AB?

(1) Angles ABC and KLM are each equal to 55 degrees.
(2) LM is 6 inches.

[Reveal] Spoiler:
Ok - my reasoning for picking answer A is based on Bunuel's explanation on similar triangles. Thanks Bunuel. But can someone please let me know how to actually calculate side AB?

Considering Statement 1

Angles ABC and KLM are each equal to 55 degrees and both triangles have angle ACB and angle KML as 90 degrees. Therefore, the third angle i.e Angle CAB and angle MKL should be equal. This is because the sum of the angles of a triangle is 180. This concludes that the two triangles are similar triangles. Therefore, the corresponding sides are in proportion. This will give us

$$\frac{AC}{KM}$$ = $$\frac{BC}{LM}$$=$$\frac{AB}{KL}$$ ---> So is the side AB will be 10??? ----------------(Here I am stuck)

Considering Statement 2

This is clearly insufficient as Just knowing LM will not give us the lenght of AB.
[Reveal] Spoiler: OA

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05 Feb 2012, 05:12
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enigma123 wrote:
The area of the right triangle ABC is 4 times greater than the area of the right triangle KLM. If the hypotenuse KL is 10 inches, what is the length of the hypotenuse AB?

(1) Angles ABC and KLM are each equal to 55 degrees.
(2) LM is 6 inches.

Below are the properties of similar triangles you might find useful for the GMAT and the solution which calculates the actual value of AB.

Properties of Similar Triangles:

• Corresponding angles are the same.
Corresponding sides are all in the same proportion.
• It is only necessary to determine that two sets of angles are identical in order to conclude that two triangles are similar; the third set will be identical because all of the angles of a triangle always sum to 180º.
• If two similar triangles have sides in the ratio $$\frac{x}{y}$$, then their areas are in the ratio $$\frac{x^2}{y^2}$$.
OR in another way: in two similar triangles, the ratio of their areas is the square of the ratio of their sides: $$\frac{AREA}{area}=\frac{SIDE^2}{side^2}$$.

Back to original question:
Attachment:

KLM Triangle.GIF [ 2.03 KiB | Viewed 7798 times ]
The area of the right triangle ABC is 4 times greater than the area of the right triangle KLM. If the hypotenuse KL is 10 inches, what is the length of the hypotenuse AB?

(1) Angles ABC and KLM are each equal to 55 degrees --> ABC and KLM are similar triangles --> $$\frac{AREA_{ABC}}{area_{KLM}}=\frac{4}{1}$$, so the sides are in ratio 2/1 --> hypotenuse KL=10 --> hypotenuse AB=2*10=20. Sufficient.

(2) LM is 6 inches --> KM=8 --> $$area_{KLM}=24$$ --> $$AREA_{ABC}=96$$. But just knowing the are of ABC is not enough to determine hypotenuse AB. For instance: legs of ABC can be 96 and 2 OR 48 and 4 and you'll get different values for hypotenuse. Not sufficient.

Hope it helps.
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Re: The area of the right triangle ABC is 4 times greater than [#permalink]

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24 May 2013, 04:21
Bumping for review and further discussion.
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Re: The area of the right triangle ABC is 4 times greater than [#permalink]

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28 Sep 2013, 00:02
Bannual, I am having trouble understanding this step in your explanation -
"• If two similar triangles have sides in the ratio \frac{x}{y}, then their areas are in the ratio \frac{x^2}{y^2}.
OR in another way: in two similar triangles, the ratio of their areas is the square of the ratio of their sides: \frac{AREA}{area}=\frac{SIDE^2}{side^2}."

When doing this problem, I chose C,

I was able to figure out that the triangles are similar but I was unfamiliar with the relationship between both hypotenuses. I thought we would need to know another side length in order to establish a ratio between one triangle and the other.

Is the relationship you described a property that we simply must memorize, or was there a way to figure it out? Can you quickly derive it? I do not understand why the lengths squared ratio is equal to the proportion of the area of both triangles.
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Re: The area of the right triangle ABC is 4 times greater than [#permalink]

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28 Sep 2013, 04:48
laserglare wrote:
Bannual, I am having trouble understanding this step in your explanation -
"• If two similar triangles have sides in the ratio \frac{x}{y}, then their areas are in the ratio \frac{x^2}{y^2}.
OR in another way: in two similar triangles, the ratio of their areas is the square of the ratio of their sides: \frac{AREA}{area}=\frac{SIDE^2}{side^2}."

When doing this problem, I chose C,

I was able to figure out that the triangles are similar but I was unfamiliar with the relationship between both hypotenuses. I thought we would need to know another side length in order to establish a ratio between one triangle and the other.

Is the relationship you described a property that we simply must memorize, or was there a way to figure it out? Can you quickly derive it? I do not understand why the lengths squared ratio is equal to the proportion of the area of both triangles.

Of course there is a way to derive this property but I recommend just to memorize it.
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Re: The area of the right triangle ABC is 4 times greater than [#permalink]

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21 Oct 2014, 07:56
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Re: The area of the right triangle ABC is 4 times greater than [#permalink]

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06 Nov 2014, 15:20
Bunuel wrote:
enigma123 wrote:
The area of the right triangle ABC is 4 times greater than the area of the right triangle KLM. If the hypotenuse KL is 10 inches, what is the length of the hypotenuse AB?

(1) Angles ABC and KLM are each equal to 55 degrees.
(2) LM is 6 inches.

Below are the properties of similar triangles you might find useful for the GMAT and the solution which calculates the actual value of AB.

Properties of Similar Triangles:

• Corresponding angles are the same.
Corresponding sides are all in the same proportion.
• It is only necessary to determine that two sets of angles are identical in order to conclude that two triangles are similar; the third set will be identical because all of the angles of a triangle always sum to 180º.
• If two similar triangles have sides in the ratio $$\frac{x}{y}$$, then their areas are in the ratio $$\frac{x^2}{y^2}$$.
OR in another way: in two similar triangles, the ratio of their areas is the square of the ratio of their sides: $$\frac{AREA}{area}=\frac{SIDE^2}{side^2}$$.

Back to original question:
Attachment:
KLM Triangle.GIF
The area of the right triangle ABC is 4 times greater than the area of the right triangle KLM. If the hypotenuse KL is 10 inches, what is the length of the hypotenuse AB?

(1) Angles ABC and KLM are each equal to 55 degrees --> ABC and KLM are similar triangles --> $$\frac{AREA_{ABC}}{area_{KLM}}=\frac{4}{1}$$, so the sides are in ratio 2/1 --> hypotenuse KL=10 --> hypotenuse AB=2*10=20. Sufficient.

(2) LM is 6 inches --> KM=8 --> $$area_{KLM}=24$$ --> $$AREA_{ABC}=96$$. But just knowing the are of ABC is not enough to determine hypotenuse AB. For instance: legs of ABC can be 96 and 2 OR 48 and 4 and you'll get different values for hypotenuse. Not sufficient.

Hope it helps.

Hi Bunuel,

Question regarding the above property.

For triangles -- if the if the sides have the ratio of x then the area will be squared. Correct? Does this imply that ALL the sides in Triangle ABC are "x times" All the sides in triangle KLM. Meaning, what happens if the sides have different ratios? base could be 2x while the height would be 3x -- what happens in a situation like that?

Thanks.
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Re: The area of the right triangle ABC is 4 times greater than [#permalink]

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07 Nov 2014, 04:34
russ9 wrote:
Bunuel wrote:
enigma123 wrote:
The area of the right triangle ABC is 4 times greater than the area of the right triangle KLM. If the hypotenuse KL is 10 inches, what is the length of the hypotenuse AB?

(1) Angles ABC and KLM are each equal to 55 degrees.
(2) LM is 6 inches.

Below are the properties of similar triangles you might find useful for the GMAT and the solution which calculates the actual value of AB.

Properties of Similar Triangles:

• Corresponding angles are the same.
Corresponding sides are all in the same proportion.
• It is only necessary to determine that two sets of angles are identical in order to conclude that two triangles are similar; the third set will be identical because all of the angles of a triangle always sum to 180º.
• If two similar triangles have sides in the ratio $$\frac{x}{y}$$, then their areas are in the ratio $$\frac{x^2}{y^2}$$.
OR in another way: in two similar triangles, the ratio of their areas is the square of the ratio of their sides: $$\frac{AREA}{area}=\frac{SIDE^2}{side^2}$$.

Back to original question:
Attachment:
KLM Triangle.GIF
The area of the right triangle ABC is 4 times greater than the area of the right triangle KLM. If the hypotenuse KL is 10 inches, what is the length of the hypotenuse AB?

(1) Angles ABC and KLM are each equal to 55 degrees --> ABC and KLM are similar triangles --> $$\frac{AREA_{ABC}}{area_{KLM}}=\frac{4}{1}$$, so the sides are in ratio 2/1 --> hypotenuse KL=10 --> hypotenuse AB=2*10=20. Sufficient.

(2) LM is 6 inches --> KM=8 --> $$area_{KLM}=24$$ --> $$AREA_{ABC}=96$$. But just knowing the are of ABC is not enough to determine hypotenuse AB. For instance: legs of ABC can be 96 and 2 OR 48 and 4 and you'll get different values for hypotenuse. Not sufficient.

Hope it helps.

Hi Bunuel,

Question regarding the above property.

For triangles -- if the if the sides have the ratio of x then the area will be squared. Correct? Does this imply that ALL the sides in Triangle ABC are "x times" All the sides in triangle KLM. Meaning, what happens if the sides have different ratios? base could be 2x while the height would be 3x -- what happens in a situation like that?

Thanks.

Again your question is not very clear... In similar triangles corresponding sides are in the same ratio.
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Re: The area of the right triangle ABC is 4 times greater than [#permalink]

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27 Mar 2015, 15:23
Bunuel wrote:
enigma123 wrote:
The area of the right triangle ABC is 4 times greater than the area of the right triangle KLM. If the hypotenuse KL is 10 inches, what is the length of the hypotenuse AB?

(1) Angles ABC and KLM are each equal to 55 degrees.
(2) LM is 6 inches.

Below are the properties of similar triangles you might find useful for the GMAT and the solution which calculates the actual value of AB.

Properties of Similar Triangles:

• Corresponding angles are the same.
Corresponding sides are all in the same proportion.
• It is only necessary to determine that two sets of angles are identical in order to conclude that two triangles are similar; the third set will be identical because all of the angles of a triangle always sum to 180º.
• If two similar triangles have sides in the ratio $$\frac{x}{y}$$, then their areas are in the ratio $$\frac{x^2}{y^2}$$.
OR in another way: in two similar triangles, the ratio of their areas is the square of the ratio of their sides: $$\frac{AREA}{area}=\frac{SIDE^2}{side^2}$$.

Back to original question:
Attachment:
KLM Triangle.GIF
The area of the right triangle ABC is 4 times greater than the area of the right triangle KLM. If the hypotenuse KL is 10 inches, what is the length of the hypotenuse AB?

(1) Angles ABC and KLM are each equal to 55 degrees --> ABC and KLM are similar triangles --> $$\frac{AREA_{ABC}}{area_{KLM}}=\frac{4}{1}$$, so the sides are in ratio 2/1 --> hypotenuse KL=10 --> hypotenuse AB=2*10=20. Sufficient.

(2) LM is 6 inches --> KM=8 --> $$area_{KLM}=24$$ --> $$AREA_{ABC}=96$$. But just knowing the are of ABC is not enough to determine hypotenuse AB. For instance: legs of ABC can be 96 and 2 OR 48 and 4 and you'll get different values for hypotenuse. Not sufficient.

Hope it helps.

Hi Bunuel,

the area of triangle should be 5 times the area of other triangle..,isn't it?
because,it mention as "4 times greater than other triangle".....
and,not as--"4 times the other triangle"

Please have a look into it..

Thanks!!
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The area of the right triangle ABC is 4 times greater than [#permalink]

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29 Dec 2015, 23:06
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution

In the figure, angle C and angle M are right angles, and KL = 10. If the area of triangle ABC is four times the area of triangle KLM, what is the length AB?
(1) Angles ABC and KLM have the same measure.
(2) LM is 6 inches.

We need to modify the original condition and the question:
First, as we can see above, KL=10. Also, taking a square root of the ratio of areas yields the ratio of lengths. Then, if the area of triangle ABC is four times the area of triangle KLM, since root of 4 is 2, we know that the length of AB is twice longer than the length of KL.
This can only be true under the assumption that two triangles have the same measure. Essentially, we can infer that the question is asking if two triangles have the same measure.
Hence, the condition 1) is sufficient and the correct answer is A.

Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
Attachments

d.jpg [ 2.89 KiB | Viewed 2158 times ]

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Re: The area of the right triangle ABC is 4 times greater than [#permalink]

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01 May 2016, 00:18
enigma123 wrote:
Attachment:
KLM Triangle.GIF
The area of the right triangle ABC is 4 times greater than the area of the right triangle KLM. If the hypotenuse KL is 10 inches, what is the length of the hypotenuse AB?

(1) Angles ABC and KLM are each equal to 55 degrees.
(2) LM is 6 inches.

So here are my 2 cents. Let's say Bigger triangle = A and smaller one = B.
Now area of A = 4 * Area of B.
and we know KL=10.

Now statement 1 says they have one angle equal and we know they both have right angles. From this we can derive that 3rd angle will also be equal. Now from this we know they are similar triangles and their sides have equal ratios.

Let say AC/KM=BC/ML=a. a is our scale factor. Now AC=a*Km and BC=a*ML
Area of KML= 1/2*KM*ML
Area of ABC = 1/2AB*BC=1/2*a*KM*a*ML= 1/2 a^2 KM&ML. You can see from here that if scale factor of sides = a then area increases by a^2.

Now lets apply this to our question. Scale factor of area = 4 then scale factor of sides will be sqrt of 4=2. Now B has KL=h=10 then AB = 2*10 = 20.

Statement2- To prove this insufficient is very easy.(Find out why)
Re: The area of the right triangle ABC is 4 times greater than   [#permalink] 01 May 2016, 00:18
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